Problem: $f(x, y) = (x\sin(y), y^2 e^y)$ $\text{curl}(f) = $
The formula for curl in two dimensions is $\text{curl}(f) = \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial Q}{\partial x} &= \dfrac{\partial}{\partial x} \left[ y^2 e^y \right] \\ \\ &= 0 \\ \\ \dfrac{\partial P}{\partial y} &= \dfrac{\partial}{\partial y} \left[ x \sin(y) \right] \\ \\ &= x \cos(y) \end{aligned}$ Therefore: $\begin{aligned} \text{curl}(f) &= 0 - x\cos(y) \\ \\ &= -x\cos(y) \end{aligned}$